# Asymptotic of $v_{n+1} = \sqrt[3]{ v_n^3 + v_n^\sqrt 7}$?

How fast is $$v_{n+1} = \sqrt[3]{ v_n^3 + v_n^\sqrt 7} ?$$ What is a good asymptotic for it ?

I know that $$v_{n+1} = \sqrt[3]{ v_n^3 + 1}$$ grows like $$v_{n+1} = \sqrt[3]{ v_1^3 + n}$$ and I assume $$v_{n+1} = \sqrt[3]{ v_n^3 + v_n^2}$$ grows more or less like $$v_1 + c n$$ for some fixed $$c$$ because $$v_{n+1} = \sqrt[3]{ v_n^3 + 3 v_n^2 + 3 v_n + 1} = v_n + 1$$ and those expressions are “close”.

But I started to really wonder about

$$v_{n+1} = \sqrt[3]{ v_n^3 + v_n^A} ?$$

For various values of $$A$$.

In particular how fast is

$$v_{n+1} = \sqrt[3]{ v_n^3 + v_n^\sqrt 7} ?$$

I assume no closed form exists.

Im not sure if using Taylor or Laurent series would help. Or maybe matrices ? The problem can be restated with Taylor , Laurent and matrices. But that does not seem to make it easier at first.

I did some numerical tests but I was not convinced of anything.

Ofcourse if one iterates $$v + q(v)$$ where $$q(v)$$ goes to 0 as $$v$$ increases we could estimate $$n$$ iterations to give $$v + \sum_i^n q(i)^{-1} q(v)$$.

For example when $$q(v) = +/- 1/v$$ this works relatively well. But not super good. ( for the negative case take v_1 > 2n )

• Why not just take $u_n=v_n^3$, when the iteration becomes $u_{n+1}=u_n+u_n^{\sqrt{7}/3}$? I don't see much point to the external cube root... Jan 5, 2020 at 22:36

I guess that you are assuming that $$v_1>0$$ (The case $$v_1=0$$ is trivial, and the $$v_n^{\sqrt{7}}$$ forces us to work with $$v_1\ge 0$$).
Now, Notice that $$v_n\to\infty$$, $$v_n$$ is increasing, $$v_{n+1}/v_n\to 1$$ and $$v_{n+1}-v_n = \frac{v_{n+1}^3-v_n^3}{v_n^2+v_n v_{n+1} +v_{n+1}^2} =\frac{v_n^{\sqrt{7}} }{v_n^2+v_n v_{n+1} +v_{n+1}^2}\to \infty$$ From this we can see that $$v_n$$ grows faster than $$cn$$.
A slightly well known technique to find the rate of growth of such sequences is mentioned in this small article, we try to find $$a$$ such that $$v_{n+1}^a-v_n^a\to L$$ where $$L$$ is a non zero real number. In this case, The only $$a$$ that works is $$3-\sqrt{7}$$ (I have found it using Mathematica, Taylor series can be used to find it). So (By stolz lemma),
$$\frac{1}{n}\sum_{k=1}^n v_{n+1}^a -v_n^a =L= \frac{3-\sqrt{7}}{3}$$ So $$v_{n+1}^a\sim L n$$. Therefore
$$v_{n} \sim \left(\frac{3-\sqrt{7}}{3} n\right)^{\frac{1}{3-\sqrt{7}}}$$
Here's a plot of $$v_n$$ in blue and the RHS in orange